# $U\subseteq V$ is $T$ invartiant $\Rightarrow$ $\left(T\,|_{U}\right)^{*}=\left(P \circ T^{*}\right)\bigl|_{U}$

I want to proove that given $T\in\mathcal{L}\left(V,V\right)$ ($V$ is a finite dimensional inner product space) and a subspace $U\subseteq V$ which is $T$ invariant that

$\left(T\,|_{U}\right)^{*}=\left(P\circ T^{*}\right)\bigl|_{U}$

Where $P$ is the projection on $U$.

My attempt: I've tried working with the inner product but but with no luck because I'm unsure how can I "defuse" all the $^*$. I've all so thought about finding some orthonormal basis of $U$ $\mathcal{B}=\left\{ f_{1},\dots,f_{k}\right\}$ but I didn't know how to express those linear transforms with the basis.

I do understand intuitively why restricting $T^{*}$ is the same as using $T^*$ first and then just projection the result to $U$ but I'm struggling writhing a correct rigorous proof.

I'd be happy to provide any additional information.

• You might want to edit your question to add that you assume $V$ to be a Hilbert space. Commented May 22, 2013 at 15:20
• @julien Thanks, but we assumed that $V$ is an inner product space...
– Scis
Commented May 22, 2013 at 15:26
• Infinite dimensional? Then you need at least $U$ to be complete for the projection onto $U$ to exist by the standard projection onto complete convex subsets of an inner product space. Commented May 22, 2013 at 15:52
• Also, try do this algebraically: $U$, the range of $P$, is exactly the vectors $v$ such that $Pv=v$. So $U$ is $T$ invariant iff $PTPv=TPv$ for every $v\in V$, i.e. $PTP=TP$. Taking the adjoint of the latter, this is in turn equivalent to $PT^*P=PT^*$, which is actually your condition, if you realize that restricting to $U$ amounts to right-multiply by $P$. Commented May 22, 2013 at 15:56
• The key point is that for an idempotent operator ($P^2=P$), the range is characterized by $\mbox{Im}\,P=\{v\in V\;;\; Pv=v\}=\mbox{Ker}\,(I-P)$. Once you have that, $U=\mbox{Im}\,P=\{Pv\;;\; v\in V\}$ is $T$ invariant iff $Tw\in \mbox{Im}\,P$ for every $w\in \mbox{Im}\, P$ iff $TPv\in\mbox{Im}\,P$ for every $v\in V$ iff $PTPv=TPv$ for every $v$ iff $PTP=TP$. And the last thing you need is that for a projection $P$ (self-adjoint idempotent), $P^*=P$. Commented May 22, 2013 at 16:30

For $u, v \in U$ we have (denoting by $(,)$ the inner product on $V$ and by $(,)_U$ its restriction to $U$): \begin{align*} (T|_Uu, v)_U &= (Tu, v)\\ &= (u, T^*v)\\ &= (u, PT^*v) + (u, (I - P)T^*v)\\ &= (u, PT^*v)\\ &= (u, (PT^*)|_Uv)_U \end{align*} So $(T|_U)^* = (PT^*)|_U$. In the forth step we used that$\mathrm{ran}(I-P) = U^\bot$ (i. e. the range of $I-P$ is the orthogonal complement of $U$, as $I-P$ is the projection onto $U^\bot$), hence $(u, (I-P)x) = 0$ for all $x \in V$.
• Can you please explain what do you mean by $ran$? this part was a bit unclear...